Number of natural number solutions I have reduced one of my problems to a seemingly simple number theoretic problem. I read a bit on diophantine equations in order to find a helpful technique, but my impression was that they always consider settings in the integers and usually polynomials of a variable, while in my case products of variables appear.
I need to know how many solutions in the natural numbers exist for
\begin{equation}
2(r*s-u*v)=d
\end{equation}
with a fixed $d$, $0\leq r,s,u,v \leq d-1$ and $r,s,u,v,d \in \mathbb{N}$.
The goal would be to find an function $f(d)$ that gives me the number of solutions for as many values of $d$ as possible. I suspect it will not be possible to find one function for all values, but any non finite subset might be interesting, the larger the better.
Does anyone know how to approach the question?
 A: Suppose $d=2p$, with $p$ an odd prime.
Firstly, I find the number of solutions when the value is a multiple of $d$.  Then, I assume these solutions are distributed normally, and use the continuous approximation to estimate the number of solutions when this sum is between $d/2$ and $3d/2$.
If $2rs$ is a multiple of $p$, then either $r=p$ or $s=p$; and $u=p$ or $u=0$ or $v=p$ or $v=0$.  There are $d-1$ solutions with $r=u=p$ (as $s=v+1$); $d-1$ with $s=u=p$; $d-1$ with $r=v=p$; $d-1$ with $s=v=p$; but $2$ with $r=s=p$ and $2$ with $u=v=p$, so $4d-8$ altogether.  If $r=p,u=0$ there are $d$ solutions $(s=1,v=$anything$)$, and $4d-2$ solutions of this type.  So $$S_1=8d-10$$ solutions when the two terms are multiples of $p$.
Now for $2rs$ not a multiple of $p$, I find solutions where $rs=uv\pmod p$.  If $u\neq0\pmod p$, then for any $r$ and $s$ there are two possible values of $v$; which gives $N=16p^2(p-1)=2d^2(d-2)$ solutions when $2rs$ and $2uv$ match $\mod p$ but are not multiples of $p$.
Since $r,s,u,v$ are not multiples of $p$, we have:$$\mu=\mathbb E(2rs-2uv)=0\\ \sigma^2=\mathbb{E}(2rs-2uv)^2=8((4d-1)d/12)^2$$
so I expect roughly $$S_2=Nd/\sigma=3\sqrt{2}(2d^2(d-2))/(4d-1)$$ solutions when $2rs$ is not a multiple of $p$.
We also need $rs$ to share no factor with $uv$.  The chance they share a prime factor $q$ is $(2q-1)^2/q^4$, so the chance they don't is $\frac{q^4-4q^2+4q-1}{q^4}$.  So $S_2$ must be reduced by the factor 
$$F = \prod_{q\text{ prime}}\frac{q^4-4q^2+4q-1}{q^4}$$
Altogether $S_1+FS_2$ solutions, if $d=2p$
